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So, we know that in traditional real numbers we have this identity.

$\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}$

Then this also work for inverses in modular arithmetic

if we have $\bar{n}\in \mathbb{Z/pZ}$ and $(\bar{n})^{-1}$ denotes the inverse of $\bar{n}$ in $\mathbb{Z/pZ}$

Then $[(\bar{n})^{-1}(\overline{n+1})^{-1}] = [(\overline{n(n+1)})^{-1}] = (\bar{n})^{-1} - (\overline{n+1})^{-1}$

The idea goes:

$[(\overline{n(n+1)})^{-1}] = [(\overline{n(n+1)})^{-1}]*\bar{1} = [(\overline{n(n+1)})^{-1}] [\overline{(n+1)-(n)}] = [(\overline{n+1}) (\overline{n(n+1)})^{-1}]-[(\bar{n}) (\overline{n(n+1)})^{-1}] = [(\overline{n+1})(\overline{(n+1)})^{-1}((\overline{n})^{-1})] - [(\overline{n})(\overline{(n)})^{-1}((\overline{n+1})^{-1}] = [(\overline{n})^{-1}] - [(\overline{n+1})^{-1}]$

Which leads me to ask if using the notation $\frac{1}{n}$ to denote inverses is completely consistent with inverses. If so, then is there some sort of homomorphism between $\mathbb{Z/pZ}$ and $\mathbb{Q}$ that simplifies the proof that we can do normal fraction rules in modular arithmetic?

Bill Dubuque
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Fernandeez nuts
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    I believe the answer is simply that as long as all the numbers involved are invertible $\pmod m$ (in particular if $m$ is prime), fractions work just like regular numbers within modular arithmetic. That particular identity isn't all that special. For instance, we can do $$\frac37+\frac5{12}=\frac{36}{84}+\frac{35}{84}=\frac{71}{84}$$ but also in $\pmod{37}$, we have $$\frac37+\frac5{12} \equiv (3)(16)+(5)(34) \equiv 33 \ ; \ (33)(84) \equiv 34 \equiv 71$$ If what you're looking for is proof... well, someone else will need to provide a link for that. – Eric Snyder Sep 06 '23 at 05:08
  • It's a nice, not too hard exercise in group theory that there are no homomorphisms between $\Bbb Z/p\Bbb Z$ and $\Bbb Q$ (in either direction) other than the constant functions that take everything to the identity. – Greg Martin Sep 06 '23 at 05:34
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    See the linked dupes. The sought map is not between $\Bbb Q$ but rather its subrings (localizations) that invert a specified subset $S$ of primes. Any equality in this subring has an (unique) image in any ring $R$ where the primes in $S$ are all units (invertibles), by mapping $,a/b,$ to $,ab^{-1},$ in $R.,$ E.g the equation in the first comment has primes $2,3,7$ in denom's so it maps to a valid equation in every ring where $2,3,7$ are units, e.g. in $,\Bbb Z/n,,$ for $n$ coprime to $2,3,7.,$ Ditto for all fraction arithmetic in this subring of $,\Bbb Q.\ \ $ – Bill Dubuque Sep 06 '23 at 07:26
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    This is the universal property of subrings of $\Bbb Q$ of this form (obtained by adjoining inverses of some elements). See localization in any textbook on commutative algebra – Bill Dubuque Sep 06 '23 at 07:26
  • @GregMartin is the proof related to the fact that $<\bar{k}> = \mathbb{Z/pZ}$ for all $\bar{k} \in \mathbb{Z/pZ}$ that is not the identity – Fernandeez nuts Sep 06 '23 at 08:04
  • That's not true: consider $p=5$ and $k=4$ for example. – Greg Martin Sep 06 '23 at 20:32
  • @GregMartin It is though because $4 \equiv 4$ (mod5), $4+4=8 \equiv 3$(mod 5), $4+4+4=12 \equiv 2$ (mod5), $4+4+4+4 = 16 \equiv 1$ (mod 5), $4+4+4+4+4 = 20 \equiv 0$ (mod 5) which gives all the residue classes. – Fernandeez nuts Sep 07 '23 at 07:13
  • The homomorphism from $(\mathbb{Z/pZ}, +)$ to $(\mathbb{Q}, +)$ must be the identity since (suppose we have a nontrivial homomorphism $f$) we must have $f(\bar{1})= q \neq 0$ , which then gives that $im(f) = {nq | n \in {1,…,n}}$ but this can only be a subgroup of Q if $q=0$ (since $$ is infinite for any nonzero $q$ but $im(f)$ only has at-most n elements. – Fernandeez nuts Sep 07 '23 at 07:49
  • However, I cannot seem to prove the other way that you also suggest. – Fernandeez nuts Sep 07 '23 at 07:56

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